Answer
\[\underline{36.9\times {{10}^{-10}}}\]
Work Step by Step
Radius of neon atom is \[\text{69 pm}\].
\[\text{1 pm}=\text{1}{{\text{0}}^{-12}}\text{ m}\]
Thus, radius of neon atom in decimeter is written as follows:
\[r=69\times {{10}^{-12}}\text{ m}\]
Assuming a spherical shape:
\[V=\frac{4}{3}\pi {{r}^{3}}\]
Thus,
\[\begin{align}
& V=\frac{4}{3}\pi {{\left( 69\times {{10}^{-12}}\text{ m} \right)}^{3}} \\
& =1.37\times {{10}^{-30}}\text{ }{{\text{m}}^{3}}
\end{align}\]
Now,
\[\text{1 L}=\text{1}{{\text{0}}^{-3}}\text{ }{{\text{m}}^{3}}\]
Thus,
\[\begin{align}
& V=1.37\times {{10}^{-30}}\,\times \,{{10}^{-3}}\text{ L} \\
& \,\,\,\,\text{=}\,1.37\times {{10}^{-33}}\,\,\text{L} \\
\end{align}\]
There are \[2.69\,\times \,{{10}^{22}}\,Ne\,atoms\,\] present in \[1\,L\]of gaseous sample. Thus, the volume occupied by \[2.69\,\times \,{{10}^{22}}\,Ne\,atoms\,\]is :
\[\begin{align}
& 2.69\,\times \,{{10}^{22}}\,\,\,\times \,\,1.37\times {{10}^{-33}}\,\,\text{L}\, \\
& =\,3.68\times {{10}^{-11}}\,\,L\, \\
\end{align}\]
The total volume of the gaseous sample is \[1\,L\].
The fraction of space occupied by the atoms is the ratio of volume occupied by the atoms to total volume of the gaseous sample. Thus,
\[\begin{align}
& \text{Fraqction of space}=\frac{3.69\times {{10}^{-11}}\,\,\text{L}}{1\text{ L}} \\
& =3.69\times {{10}^{-11}}\,\approx \,\,36.9\times {{10}^{-10}}
\end{align}\]
The fraction of the space occupied by the atoms is \[\underline{\,\,36.9\times {{10}^{-10}}}\].
